a.

\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{2}\)

\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)

\(-\frac{5}{6}\times x=\frac{5}{2}\)

\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)

\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)

\(x=-3\)

b.

\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)

\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)

\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)

\(3x-\frac{37}{10}=-\frac{19}{2}\)

\(3x=-\frac{19}{2}+\frac{37}{10}\)

\(3x=\frac{-95+37}{10}\)

\(3x=-\frac{58}{10}\)

\(3x=-\frac{29}{5}\)

\(x=-\frac{29}{5}\div3\)

\(x=-\frac{29}{5}\times\frac{1}{3}\)

\(x=-\frac{29}{15}\)

c.

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21-16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}\times\frac{4}{3}\)

\(x=\frac{5}{6}\)

d.

\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)

\(-\frac{2}{3}\times x=\frac{3}{10}-\frac{1}{5}\)

\(-\frac{2}{3}\times x=\frac{3-2}{10}\)

\(-\frac{2}{3}\times x=\frac{1}{10}\)

\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)

\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)

\(x=-\frac{3}{20}\)

e.

\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)

\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)

\(\left|x\right|=\frac{20+9}{12}\)

\(\left|x\right|=\frac{29}{12}\)

\(x=\pm\frac{29}{12}\)

Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)

f.

\(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)

\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)

\(\left|2x-\frac{1}{3}\right|=\frac{6-5}{6}\)

\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)

\(2x-\frac{1}{3}=\pm\frac{1}{6}\)

                \(2x=\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{1+2}{6}\)

                \(2x=\frac{3}{6}\)

                \(2x=\frac{1}{2}\)

                  \(x=\frac{1}{2}\div2\)

                  \(x=\frac{1}{2}\times\frac{1}{2}\)

                  \(x=\frac{1}{4}\)

                \(2x=-\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{-1+2}{6}\)

                \(2x=\frac{1}{6}\)

                 \(x=\frac{1}{6}\div2\)

                 \(x=\frac{1}{6}\times\frac{1}{2}\)

                 \(x=\frac{1}{12}\)

Vậy x = 1/4 hoặc x = 1/12.

Chúc bạn học tốt

Sorry nha, mik chép lộn đềLàm lại câu a nha

a.

\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{12}\)

\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{12}\)

\(-\frac{5}{6}\times x=\frac{5}{12}\)

\(x=\frac{5}{12}\div\left(-\frac{5}{6}\right)\)

\(x=\frac{5}{12}\times\left(-\frac{6}{5}\right)\)

\(x=-\frac{1}{2}\)

Chúc bạn học tốt

\(a,\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\frac{4}{6}x-\frac{9}{6}x=\frac{5}{12}\)

\(\frac{-5}{6}x=\frac{5}{12}\)

\(x=\frac{5}{12}:\frac{-5}{6}\)

\(x=\frac{-1}{2}\)

\(b,\frac{2}{5}+\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}\)

\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{4}{10}\)

\(3x-3,7=\frac{-57}{10}:\frac{3}{5}\)

\(3x=\frac{-19}{2}+\frac{37}{10}\)

\(3x=\frac{-95}{10\text{​​}\text{​​}\text{​​}}+\frac{37}{10}\)

\(x=\frac{-29}{5}:3\)

\(x=\frac{-29}{15}\)

Câu a \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)

g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)

Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

i) \(\frac{6}{2}=\frac{-5+x}{15}\)

\(\Leftrightarrow3=\frac{x-5}{15}\)

\(\Leftrightarrow x-5=15.3\)

\(\Leftrightarrow x-5=45\)

\(\Leftrightarrow x=45+5\)

\(\Leftrightarrow x=50\)

a. \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(\Rightarrow2+\frac{3}{4}x=\frac{7}{9}:\frac{8}{27}\)

\(\Rightarrow2+\frac{3}{4}x=\frac{21}{8}\)

\(\Rightarrow\frac{3}{4}x=\frac{21}{8}-2\)

\(\Rightarrow\frac{3}{4}x=\frac{5}{8}\)

\(\Rightarrow x=\frac{5}{8}:\frac{3}{4}\)

\(\Rightarrow x=\frac{5}{6}\)

b. \(\frac{-2}{3}x+\frac{1}{5}=\frac{3}{10}\)

\(\Rightarrow\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}\)

\(\Rightarrow\frac{-2}{3}x=\frac{1}{10}\)

\(\Rightarrow x=\frac{1}{10}:\left(\frac{-2}{3}\right)\)

\(\Rightarrow x=\frac{-3}{20}\)

c. \(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)

\(\Rightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)

\(\Rightarrow\left|x\right|=\frac{29}{12}\)

\(\Rightarrow x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)

yutyugubhujyikiu

\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)

\(=\frac{7}{2}-2\)

\(=\frac{7}{2}-\frac{4}{2}\)

\(=\frac{3}{2}\)

\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)

\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)

\(=\frac{3}{7}.\left(2-9\right)\)

\(=\frac{3}{7}.\left(-7\right)\)

\(=-3\)

\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )

1 

a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)

= \(3\cdot25:\frac{5}{4}\)

= \(3\cdot\left(25:\frac{5}{4}\right)\)

=\(3\cdot20\)

=60

b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)

=\(\frac{3}{7}\cdot\left(-7\right)\)

=\(-3\)

c) = 

\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)

=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)

=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)

=> \(-\frac{3}{4}+\left(-2x\right)=-2\)

=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)

=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)

Vậy \(x\in\left\{\frac{5}{8}\right\}\)

\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)

=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)

=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)

=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)

=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)

Vậy \(x\in\left\{-\frac{39}{40}\right\}\)

\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)

=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)

=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)

( chiệt tiêu )

=> \(5x-6x+26=-14-7x\)

=> \(-x+26=-14-7x\)

=> \(-x+7x=-14-26\)

=> \(6x=-40\)

=> \(x=-40:6=\frac{20}{3}\)

Vậy \(x\in\left\{\frac{20}{3}\right\}\)

\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)

=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)

( chiệt tiêu )

=> \(2\left(2x-3\right)-9=5-3x-2\)

=> \(4x-6-9=3-3x\)

=> \(4x-15=3-3x\)

=> \(4x+3x=3+15\)

=> \(7x=18\)

=> \(x=18:7=\frac{18}{7}\)

Vậy \(x\in\left\{\frac{18}{7}\right\}\)

\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)

ĐKXĐ : \(x\ne0\)

=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)

=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)

=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)

=> \(\frac{32}{3x}=\frac{1}{4}\)

=> \(3x=32.4:1=128\)

=> \(x=128:3=\frac{128}{3}\)

Vậy \(x\in\left\{\frac{128}{3}\right\}\)

\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)

ĐKXĐ :\(x\ne1;\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)

=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)

=> \(\frac{26+5-2}{2\left(x-1\right)}\)

=> \(\frac{29}{2\left(x-1\right)}\)

\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)

=> \(x=\frac{19}{10}:2=\frac{19}{20}\)

Vậy \(x\in\left\{\frac{19}{20}\right\}\)

\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)

=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)

=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)

=> \(x=\frac{1}{2}:2=\frac{1}{4}\)

Vậy \(x\in\left\{\frac{1}{4}\right\}\)